help-36

[quasi lotus]

for this one I got h(10.5) = (-1/7)(10.5)^2+6(10.5)

is equal to 47.25

[olive topaz]

There's two values of w

Which is 10.5 and 31.5

[quasi lotus]

ohh

so its 32?

[olive topaz]

Nope

Let assume the straight line inside the graph is at a height of 47.25

U can see the line intercept at two point right

When w=10.5, 31.5

[quasi lotus]

yeah

[olive topaz]

Then the line is the width right

[quasi lotus]

ohh so 21?

[olive topaz]

Ye

[quasi lotus]

thank you so much

[olive topaz]

Ur welcome

[quasi lotus]

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[desert juniper]

someone please consider helping me i got the first bit of z1 and z2

i dont get how to find the acute angle

is it both of the angles combined?

or angle between?

but that would be too big

[silver maple]

Use the theorem $a.b = |a|*|b|*cos(theta)$

[soft zealot]

team132

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@desert juniper Has your question been resolved?

[desert juniper]

i dontthink that works

@silver maple i think that angle would be too big no?

its acutte

acute

[silver maple]

It will work. Do you know what a dot product is

If you want an acute angle you could use 180 -angle

Another way would be consider lines like y = mx +b. Then using tan(theta1 -theta2) = ...

[desert juniper]

iidont know sorry

im in year 12

[silver maple]

Google dot product first, youtube videos perhaps

[desert juniper]

ahh ic

[silver maple]

Then apply the theorem I mention above

[desert juniper]

but basically

the scalar product

is like

use for tht

[silver maple]

scalara product is the dot product

[desert juniper]

but it will be in like

complex form right?

i have an answer but im not sure of it

could it be that?

i simplified from scalars

because it would be -28-4i=20root2cos0

[silver maple]

If the angle is too big maybe multiply by -1

[desert juniper]

i think its q

like

what could q value be

[silver maple]

E.g. cos^-1(-1) = 180 , cos^-1(1) =0

[silver maple]

Use the formula I mention above to verify

[desert juniper]

i think it works

is there anyway you could check the answer with me?

because

i dont have any

and am unsure

could it q= -7/10 +i/10

[desert juniper]

.lose

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[elfin quartz]

for this question, why did they use 1 and 1.5 for part a and 1.5 and 2 for part b

[valid rapids]

so the question is getting you to split the area under the curve into two rectangles

and for a it wants you to take the height of the rectangle as the y value when the inner/left corner of the rectangle touches the curve

if it’s between x = 1 and x = 2

the rectangles are from x = 1 to x = 1.5 and x = 1.5 to x = 2 if split evenly

so for a you are using the left corners so x = 1 and x = 1.5

and for b you are using the right corners

so x = 1.5 and x = 2

[elfin quartz]

ooh okay thank you but what if its between 1 and 3?

or like 1 and 4

would it just be 1 and 1.5, then 3.5 and 4?

[valid rapids]

you need to divide the area into rectangles

so find the average of the two x values and that will be the right side of the first rectangle and the left side of the second rectangle

for 1 and 4

it would be 1 and 2.5 then 2.5 and 4

if you are using two rectangles

[elfin quartz]

i think i get it now, thank you

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[mighty osprey]

How many simplified fractions with denominator 1980 that is less than 1 are there?
Idk the approach to this question but through some coding i found the answer is 480. Can anyone help

[lilac wing]

This means that the highest common factor between the numerator which is unknown and the denominator which is 1980 is one
This’ll happen if is a multiple of an odd number which are not multiples of 3,5 or 11
Solving this using permutations u can get the solution

[mighty osprey]

what are permutations

[lilac wing]

Arrangement of objects

[lilac wing]

All prime numbers other than 3,5 and 11 will work
One will also work
And odd numbers which ain’t multiples of 3,5,11 will work

[mighty osprey]

i dont think i can count the number of prime numbers to 1980 in a test lol

[lilac wing]

Ok

Use

[mighty osprey]

do i start buy eliminating odds that are multiple of 5,11?

[lilac wing]

Sets and relations

[lilac wing]

Yes

Draw set containing all 3 multiples
Set containing 5 multiples
Set containing 11 multiples
Find the union
n(3m) + n(5m) + n(11m) - n(3,11m) - n(3,5m) - n(5,11m) + n(3,5,11m)

Where n(Km) refers to no of k multiples till 1980

[lilac wing]

Find the value of this
Subtract it 1980

[mighty osprey]

so 1980/3 = 660 multiple3s
1980/(35) = 132
1980/(311) = 60
660-132-60 = 468

[lilac wing]

And easy way to find it using GIF

[lilac wing]

35?

Lemme see if I get the answer

Wait

I made a critical mistake

It has to be only odd multiples of 5

And it has to be only odd multiples of 3 and odd multiples of 11

Forgive me for my carelessness

[mighty osprey]

so 3,5,9,11,15,...?

[final saddle]

@mighty osprey Has your question been resolved?

[hollow timber]

anyone wanna solve 5000 problems in the next 50hr

[mighty osprey]

using sets i got 980

no

yeah yours is actually correct

i see

i got 960 for 3,5,11 without 2

all the multiples of 2 without 3,5,11 is just 960/2

were now left with 480

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[strange swan]

Is this right? Why is it 5/72π in the answer sheet

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[marble agate]

dont troll pls

[final tangle]

the level of difficulty requires the aid of <@&268886789983436800>

[marble agate]

fr

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[warm python]

asked this earlier, but say I have a function $\sin\left(x\right)+\cos\left(x+\phi\right)$, how do I find its maxima and minima using phasors ?

[soft zealot]

Why am. I here

[warm python]

I mean , phasors are just vectors, right?

so can can I exploit that somehow?

Myabe using the cosine rule?

but that wouldn't help much, would it?

as the functions are still variables

hmm, so that would be

$\sqrt{\sin^2\left(x\right)+\cos^2\left(x+\phi\right)+2\sin\left(x\right)\cos\left(x+\phi\right)\cos\left(\phi\right)}$

[soft zealot]

Why am. I here

[warm python]

which doesn't help much

I mean I guess I could solve it at x=0

but that's just sus

I mean, I guess I could evauvate it at x=0 as the phase difference is constant

here's some additional context https://en.wikipedia.org/wiki/Phasor

<@&286206848099549185>

[prisma anvil]

Mother father gentleman

[warm python]

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[bronze cloud]

help me 🥺

[hollow iron]

Write down what f(g(x)) is first

[bronze cloud]

sqrt 72-x^2+x

[warm ether]

ya sure?

[bronze cloud]

+x instead of -x?

now what?

[warm ether]

you can look directly at fog
or you can look at the domain of f alone first

[bronze cloud]

would the function of f be 72?

[warm ether]

that doesnt mean anything

rephrase

[bronze cloud]

would the x in the equation of f(x)=72?

[warm ether]

for what reason

[bronze cloud]

72-72=0

[warm ether]

theres nothing wrong with x=72, thats in the domain of f

[bronze cloud]

so what amm i supposed to do?

[warm ether]

lets go back to the parent function then sqrt(x)

whats the domain of that

[bronze cloud]

of f(x)?

[warm ether]

of sqrt(x)

[bronze cloud]

sqrt 72-x

[warm ether]

im asking you what the domain of sqrt(x) would be

[bronze cloud]

i do not know

[warm ether]

alright, can you take the root of -5

[bronze cloud]

no?

[warm ether]

-1?

[bronze cloud]

huh

[warm ether]

can you do sqrt(-1) either

[bronze cloud]

either what?

[warm ether]

does sqrt(-1) have a real value

[bronze cloud]

nooo

[warm ether]

so you agree for sqrt(x), x cant be negative

[bronze cloud]

that makes sense

[warm ether]

that means the domain of sqrt(x) is x>=0

[bronze cloud]

ok

[warm ether]

now, f=sqrt(72-x)

[warm ether]

with this in mind, what restriction is 72-x under

[bronze cloud]

x needs to be greater than 0

[warm ether]

does it?

what if x was -2

sqrt(74) isnt a problem

[bronze cloud]

72-x needs to be greater than 0?

[warm ether]

or equal, yeah

72-x>=0 72>x

thats fs domain

we subbed g into f

[warm ether]

so we can replace this x with g

72>g(x)

72>x^2-x

[bronze cloud]

ok

what do i do after that?

[warm ether]

solve it

[bronze cloud]

how with x^2?

do i square root everything?

[warm ether]

doesnt work like that

treat it in much the same way youd solve ax^2+bx+c=0
from x^2-x-72<0

[bronze cloud]

quadratic eqation?

[warm ether]

if you need to

[bronze cloud]

Is this the right formula❓

[warm ether]

brackets matter, but also no

denominator is wonky

and signs are wonky

[bronze cloud]

how?

[warm ether]

what is the quadratic formula

[bronze cloud]

-b+ or - sqrt b^2 -4(a)(c) all divided by 2a

[warm ether]

and what are a,b and c here

[bronze cloud]

a=72 b=-1^2and x=1

[warm ether]

no

[bronze cloud]

is b supposed to be squared

[warm ether]

no, im just asking what a,b and c are

from ax^2+bx+c

and by extension in the quadratic formula

[bronze cloud]

is a=72?

[warm ether]

no

there is not a 72 in front of the x^2

[bronze cloud]

so a=-1?

[warm ether]

no

a=1

[bronze cloud]

is it cuz its sqared?

[warm ether]

its not squared

x^2-x-72

there is just a 1 there

there was never a -1

[bronze cloud]

but f(x) is 72-x

and i replaced x with g(x)

so it became 72-x^2+x

[warm ether]

well, its 72-x^2+x>=0 which is the same as 0>=x^2-x-72

but if you want to do it that way, then fine

[bronze cloud]

so would with my way a=-1 b=1 and c=72?

[warm ether]

it would

[bronze cloud]

Is this the correct formula❓

i solved it and got -8,9

@warm ether

woulld it be brackets or parenthises?

i got it right byebye

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[distant cloud]

Please how to resolve this inesuation, because y can’t use ln in this situation no ?

[distant cloud]

Inequation *

I’m stuck

[solar glade]

It is -(m+1)/2? Or (-m+1)/2?

You can take common factor

(1-sqrt(e))(200e^(-(m+1))/2)

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[distant cloud]

It’s -((n+1)/2)

I don’t understand your commun factor

sqrt(e)(200e^(-(m+1))/2) = 200e^(-(m/2)) ???

Ok i understand but Ist (sqrt(e)-1) no ?

Its*

[solar glade]

-m-1

It is 1-sqrte

[distant cloud]

Look

I’m wrong ?

[solar glade]

Yes, try factoring correctly

The result of factoring should be (1-sqrt)f

Being f =e^blabla

[distant cloud]

Ok anyway after I can’t put it in fonction ln because it’s negative ?

[solar glade]

It is not negative

You will use that

To divide 5

And ln the e^etc

Remember u have

A 200

U will use that too

So u will have

Negative times negative in rhs

= positive

[distant cloud]

1-sqrte it’s negative and exp(bla-bla-bla) it’s positive

So it’s negative

I think it’s sqrt(e)-1 but I’m just a good 18 years old

If you sure I believe you

[solar glade]

When u divide

Divide between -200

Wait

I made a typo

Now i see why u dodnt udnerstand

[distant cloud]

Yes it’s 200

[solar glade]

It is -200e^

[distant cloud]

But I understand

You help me

[solar glade]

Thats why didnt make sense to u

[distant cloud]

Thx

Ohhhhh

Let me check

[distant cloud]

Here it’s -200 ?

[solar glade]

(1-sqrte)(-200e^(-(m+1)/2))

[distant cloud]

Ok ok

Thx now that will work

[solar glade]

Yep

[distant cloud]

Thx

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[mighty sentinel]

+-X +-Y +-Z = A
Say I don’t know the signage of X Y and Z, but I do know that a solution exists to make it equal to A. How would I solve this?

[cursive bough]

we would need way more info here. what is A? are x, y, z integers, real numbers, etc.

are they being added or multiplied?

[mighty sentinel]

They are being summed together, and they are Positive Integers

[sturdy cypress]

the largest one is plus, the second largest is minus

only 2 cases to check

wait what

there aren't any cases

it's just +−−

where largest is plus

oh, i;m thinking of A =0 i see

[mighty sentinel]

Or as an example, (positive/negative 150) + (positive/negative 125) + (positive/negative 200) = 175

which is -150 + 125 + 200, but I was curious how one could get to that answer without guessing

[sturdy cypress]

guessing seems very effective

we guess 150 is positive
then we get
150 ± 125 ± 200 = 175
± 125 ± 200 = 25
so we were wrong, it's negative
−150 ± 125 ± 200 = 175
± 125 ± 200 = 325

you guess any number and eliminate it entirely, and it immediately becomes trivial

[mighty sentinel]

What about with 41 numbers, I would have to do roughly 1 trillion guesses or (2^(41-1))

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[sturdy cypress]

you just avoid guessing things that are too large

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[hushed dust]

Hello, how would you solve this?

[hushed dust]

I tried to do this but I don't know how to continue

I have to simplify the trigonometric expression

The answer is

[gritty rune]

Just asking, what does SEN represents?

[void valley]

sine

[gritty rune]

Oh

[void valley]

it's just notation used for example in Spain

[gritty rune]

That makes sense

Notice that:
$$\sin \Big(15^{\circ} + \frac{\alpha}{4}}\Big)=\cos \Big(75^{\circ+\frac{\alpha}{4}})$$

[soft zealot]

fAaAtDoOoG
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

Modus
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[gritty rune]

Notice that:
$$\sin \Big(15^{\circ} + \frac{\alpha}{4}}\Big)=\cos \Big(75^{\circ+\frac{\alpha}{4}})\Big$$

[soft zealot]

fAaAtDoOoG
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[gritty rune]

Nvm

I gave up

[void valley]

Delete it

[hushed dust]

yes

[void valley]

and then

you can use e.g. sin(2x) formula

[soft zealot]

Modus

[void valley]

hence in the denom you'll get:

[soft zealot]

Modus

[void valley]

and that's nice

because now you can do:

[soft zealot]

Modus

[void valley]

you see where is it going to?

[hushed dust]

not entirely

[void valley]

good

now

you know that sen(180 deg - x) = sen(x)?

[hushed dust]

yes

[void valley]

so the bottom thing is

sen(180 - (150 - a/2)) = sen(180 - 150 + a/2) = sen(30 + a/2)

do you agree

[hushed dust]

hmm, why not 30°?

[void valley]

ye 30 deg

[hushed dust]

at the beginning

[void valley]

sen(180 - x) = sen(x)

let x = 150 - a/2

[hushed dust]

ah

[void valley]

and then it cancels nicely

[hushed dust]

it was the same

thank you

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[normal juniper]

Can someone pls explain why this is true

[normal juniper]

the original equation is z^5 + 1 = 0

and they want you to factor it

oh wait nvm

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[tranquil wyvern]

Can someone help me with this one? I tried substitution but the number got really messy and i think i was goingin the wrong direction

[vital surge]

which substitution?

the one you want is ||v = y/x||

[tranquil wyvern]

oh

ok let me try that

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[wintry cedar]

Is this right?

[wintry cedar]

Original question.

The answers i'm getting are not answer choices on the question.

[vital surge]

that's a pretty good point

[final tangle]

your work looks fine
seems they forgot the cube roots

[wintry cedar]

Thanks @vital surge @final tangle

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[young verge]

How do i solve this

Im having trouble specifically with the square root part of the integral

[iron dagger]

you can attempt a trig substitution

[young verge]

Can u walk me through that

[iron dagger]

maybe inside the radical you can factor out a 3/4 to have $\sqrt{\frac{3}{4}\cdot(\frac{24}{3}+y^2)}$

[soft zealot]

Triaxyz

[iron dagger]

then pull out the $\sqrt{\frac{3}{4}}$ to leave the main radical in the form $\sqrt{a^2 + x^2}$

[soft zealot]

Triaxyz

[young verge]

Wait hows it still 24/3

And why 24/3?

[iron dagger]

multiplying that quantity by 3/4 gives 6.

I just factored out 3/4 to make the constant on y^2 equal 1

[young verge]

Oh i see

[iron dagger]

then from there you can use trig

[young verge]

Should i make it 24/4 instead tho?

so that it can combine the fraction

[iron dagger]

thats not how factoring works

[young verge]

But if i turn the 6 into 24/4 That would be equivalent wouldnt it

[iron dagger]

that would be equivalent if you kept it as 24/4 + 3/4*y^2

but if you went ahead with the factoring like I just did then it wouldn't be valid

[young verge]

Wait but what if i do

(24 + 3y^2 )/4

And then i take out 3/4

Would that be valid as well?

[iron dagger]

that would be the same thing I did, yes

[young verge]

oh sorry

And then i would use tan trig sub right

[iron dagger]

yes

both terms are positive so tan

[young verge]

Okay do u mind checking my work so far

,rotate

[iron dagger]

looks right so far

[young verge]

How should i integrate the sec^3?

Would i split them up

into sec^2 * sec

[iron dagger]

maybe int by parts since sec^2 can be integrated to tan

[young verge]

and then integration by parts?

Which would i set as my u and dv tho

Since theyre both trig

Does it matter?

[iron dagger]

I believe if they're both of the same priority then you just choose whichever one turns out to be more convenient

dont quote me on that though cause I'm not 100% sure if this is the case

[young verge]

Which would be more convenient tho

[iron dagger]

.

[young verge]

Cause im not really sure how to take the anti-derivative of sec

[iron dagger]

you already got half the procedure

split into sec and sec^2

what is the integral of sec^2?

[young verge]

tan

[iron dagger]

yes

if integrating that gives you a clean answer, would you set it as u or dv?

[young verge]

u as sec^2

dv as sec

[iron dagger]

no

u needs to be differentiated to find du

[young verge]

Oh dv as sec^2

[iron dagger]

dv needs to be?

[young verge]

Mbmb

[young verge]

sec^2

[iron dagger]

integrated but yes

[young verge]

yea that makes sense

Ok gimme a sec

[final saddle]

@young verge Has your question been resolved?

[final saddle]

@young verge Has your question been resolved?

[young verge]

Does this seem right took me a while to get it

[final saddle]

@young verge Has your question been resolved?

[final saddle]

@young verge Has your question been resolved?

[final saddle]

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[gleaming bluff]

Can someone help me with this

[gleaming bluff]

So i used graphing calculator to draw it okay

And then it asks me to find the local max value

I cant find it bx the graph looks weird

[austere heart]

Have you done derivatives

[stone wagon]

,rccw

[austere heart]

You could derive it and make a line and check the signs

[gleaming bluff]

Idk how to derive

i may be knowing this from school but maybe in another language

[austere heart]

The questions seems like calculus stuff to me tho

[gleaming bluff]

Could you explain it

[gleaming bluff]

This is precalc

[austere heart]

You find the slope at that point

[gleaming bluff]

I havent seen calc yet

[gleaming bluff]

Alr

[austere heart]

You've done it yes?

[gleaming bluff]

There is no specific point thats the case

[gleaming bluff]

The graph looks weird and it tells me to find max

[austere heart]

Don't worry just derive the whole function

[gleaming bluff]

o no i dont know any of this

[austere heart]

Do you know the power rule?

oh

[gleaming bluff]

nope im sorry

havent seen calc yet

[austere heart]

What seriously? It says where it's "increasing / decreasing" that sounds like calculus to me

Ok do you know limits?

[gleaming bluff]

i dont know anything about calc and i just started math

[austere heart]

These?

hm...

[gleaming bluff]

i wanted to start from pre calc

im grade 12 but havent seen grade 11 and i dont go to school soo

[austere heart]

Are you sure this is from pre calc tho?

This question doesn't seem like pre calc

[gleaming bluff]

James stewart’s precalc book

[austere heart]

So they want you to use algebra?

[gleaming bluff]

i suppose yes since the book never mentioned about calc ( limits etc ) until now

[austere heart]

Have you done like geometry, and algebra by yourself?

[gleaming bluff]

Yea

i completed algebra 2 first now im doing pre calc

[austere heart]

Which unit is that question from

[gleaming bluff]

Chapter 2 functions

[austere heart]

Which question

I have the pdf

[gleaming bluff]

Page 236 question 22

[austere heart]

Man... this is so weird

Can you show me the graph

I can't see the front most value of the function

[gleaming bluff]

i used desmos to graph it

lemme send a pic

[austere heart]

It's a little blurred

[gleaming bluff]

Oh sorry

Wait

The graph

The question

[austere heart]

Alr we have the graph and no calculus, that's all we need

I guess we can work wit hthis

[gleaming bluff]

Aight

[austere heart]

so I think A is just wanting you to do it with a graphing calc and zoom fit

[gleaming bluff]

yep thats done

[austere heart]

mhm now onto c

[gleaming bluff]

I found local min

[austere heart]

So you notice what it's saying

what is the local min

[gleaming bluff]

-1.61

X = -1.61
Y = -27.18

should i send u the answer sheet

For this question

[austere heart]

Umm

It said "local" min

There's a difference between absolute min and local min

[gleaming bluff]

whats the difference

[austere heart]

See, if you look here

We can see that -1.61 is the LOWEST value of the WHOLE function. Would you agree?

[gleaming bluff]

yeah

[austere heart]

So we would say that's actually the absolute min because it's the lowest value of the whole function

Now the relative min, also known as "local" min

Is also another low value, but it's not the LOWEST of the whole function

What value do you think it's also another low

[gleaming bluff]

1.429

[austere heart]

boom so that's the local min

[gleaming bluff]

So its the second one

always

second lowest one

[austere heart]

no no no not "always"

yeah second lowets one

[gleaming bluff]

oh

[austere heart]

but that's the thing

you can have multiple lowest ones

They'd all be "local mins"

But there would be ONE aboslute min

[gleaming bluff]

Which is the lowest

Am i right

[austere heart]

Yeah, the lowest of the whole function

[gleaming bluff]

oh okay i got it

[austere heart]

Nice

So you answered one part of it, what about the other part?

[gleaming bluff]

You see a little peak there in the picture

Is that the local max

[austere heart]

uh huh

yup

that's the local max, because of the same reasoning

[gleaming bluff]

is it also the absolute max

At the same time

[austere heart]

technically we should use calc to prove it's a local max but Ig the book hasn't taught it yet

[gleaming bluff]

yep

[austere heart]

that's a good question

The answer, in this case, is actually no

Cuz if you look at the function

It keeps going up

[gleaming bluff]

yeah

that was the part that confused me

[austere heart]

after 2.022 and before -2.357

It keeps going up

[gleaming bluff]

yep

[austere heart]

so there is no "absolute max"

Because it keeps going up

[gleaming bluff]

ohh

[austere heart]

but there's a local max down there

[gleaming bluff]

so if the same happened downwards there would be no absolute minimum as well?

[austere heart]

uh huh

[gleaming bluff]

for other functions

[austere heart]

yup

[gleaming bluff]

alright

Now i get it

btw ur rlly good at explaining

Thank u

[austere heart]

Np

let's look at d

What do you think is the range? This is a polynomial function

And it seems like there isn't any number that you can't put in x

so what do you think is the range though?

The domain is all real numbers

[gleaming bluff]

well since it’s continuing to increase i cant assume something

wait do i know range wrongly

Is this incorrect?

[austere heart]

Hm, you sort of have it wrong here

[gleaming bluff]

should it be otherwise?

[austere heart]

no this is correct

[gleaming bluff]

Oh

[austere heart]

domain are the inputs, range is the output

But, if you think about it, we said that it keeps going up right

Like this part, it just keeps going up

[gleaming bluff]

Yep can we say iy starts here and goes on forever

[austere heart]

Almost there

[gleaming bluff]

It starts at -1.61

[austere heart]

What's the lowest value that it starts at

[austere heart]

Close close

[gleaming bluff]

This one

[austere heart]

We're talking about range

[gleaming bluff]

oh so -27.182

[austere heart]

Which means you want the y value

Yeah

[gleaming bluff]

and + ♾️

?

[austere heart]

The lowest value of the whole function is -27.182, so it starts at -27.182

[gleaming bluff]

yep

[austere heart]

You've taught yourself well. Nice job

The range is in fact $[-27.182, \infty)$

[gleaming bluff]

Thank youu

[soft zealot]

Kai

[austere heart]

this is the official way of writing it on the test

[gleaming bluff]

oh okay

[austere heart]

Btw make sure you always use () when adressing infinity, as infinity is not "bounded"

[gleaming bluff]

🫡 got it

[austere heart]

onto e

Where do you think the function in increasing

[austere heart]

We read a graph from left to right

[gleaming bluff]

This part from the lowest to this little peak

[austere heart]

ok, that's one part!

where else is it increasing

[gleaming bluff]

And this to infinity

The part after the second lowest value

To infinity

[austere heart]

yeah, the lowest point to that local max is increasing AND from local min to infinity

both of those domains is when the function is increasing

You got it 👍

[gleaming bluff]

thank u

[austere heart]

Can you write that in official notation?

The parts where it's increasing

[gleaming bluff]

i’ll try so its [1.429, + inf )

im putting the x value right?

[austere heart]

looks correct

[gleaming bluff]

Alr and the other one is [-1.61, 0.181]

[austere heart]

almost there

here's the thing

at THAT POINT, when it's x = -1.61

The value isn't actually increasing. It's slope is zero

[gleaming bluff]

oh

Yeah ur right

[austere heart]

And that's how you'd know it's a minimum in calculus actually

[gleaming bluff]

so we dont include it

[austere heart]

You'd find out that it's slope is zero, so you'd think it's an important point on the graph. We call this a "critical" point

Yeah, don't include it but include everything after it

same with 0.181

[gleaming bluff]

alright

[austere heart]

Don't include it, but everything before it

[gleaming bluff]

But when somethinf starts ay 0.181 i dont include it as well

For the next ones

am i correct

[austere heart]

if you want to have decreasing yes

[gleaming bluff]

Yeah thats what im talkin abt

[austere heart]

you'd need (0.181, 1.429)

since it's not decreasing or increasing at the specific point

Yeah, you got it

[gleaming bluff]

I shouldn’t be including 1.429 as well right

[austere heart]

nope

[gleaming bluff]

Okayy

[austere heart]

Nice job

Keep learning 👍

[gleaming bluff]

Thank you

[austere heart]

I'd also recommend khan academy tbh. It's pretty good for precalc and calc

Anything else needs work on

[gleaming bluff]

I was actually gonna use it ( i used it for alg 2 )

[austere heart]

Yeah I'd totally recommend it

[gleaming bluff]

But someone said bc i need precalc at uni level itd be better to move from books

Khan academy’s explanation is great this

Though

[austere heart]

Use both

[gleaming bluff]

Especially to learn new things in my case

[gleaming bluff]

Alright imma add it to my courses

again thank you for the help I appreciate it

[austere heart]

Don't mention it

👍

[gleaming bluff]

I need to close this channel so goodbye

.close

[final saddle]

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[lilac bison]

4^x + x = 460

[pearl fable]

what does bro want

[stone wagon]

this can't be solved exactly

[lilac bison]

It can tho

[stone wagon]

or at least within elementary functions

[pearl fable]

he is trolling

[lilac bison]

I saw it on a YouTube thumbnail

[final saddle]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[lilac bison]

Can't find the video

[stone wagon]

why not watch it then

[stone wagon]

lol

well, here's your answer

[lilac bison]

Can't find it

[compact breach]

I figured this will be solved using lambert W

[stone wagon]

yes

[pearl fable]

ofc

[compact breach]

I saw a lot of those vids

[pearl fable]

bro is chatgpt

[stone wagon]

lmao yes

[pearl fable]

lil bro casually dropped a prompt and thought we wouldn't notice

[final saddle]

@lilac bison Has your question been resolved?

[lilac bison]

Has it bot

Has it?

You should learn to read the room

[pearl fable]

haha?

[lilac bison]

You are laughing the bot is clueless and you are laughing

[final saddle]

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[warm root]

I think sin^2v can be written to 1-cos^2v

[unique aspen]

it can, you're right, but that doesn't help too much here

[warm root]

how does one solve it easy, please help

[unique aspen]

do you know double-angle identities

[warm root]

cos2v=cos^2v-sin^2v

not 100% sure about it

[unique aspen]

right, that's it

so the key is to go a bit further

so cos^2(v) - sin^2(v) = cos^2(v) - (1 - cos^2(v)) = 2cos^2(v) - 1

so cos(2v) = 2cos^2(v) - 1

so you can now use that to undo it

or similarly cos^2(v) - sin^2(v) = (1 - sin^2(v)) - sin^2(v) = 1 - 2sin^2(v)

so cos(2v) = 1 - 2sin^2(v)

[warm root]

u type fast

[unique aspen]

sure

practice

[warm root]

I need some minutes to process this information

I bet u can do 3rd intergrals in your head xd

[unique aspen]

haha no

[warm root]

ok I get how you came to 1-2sin^2v

now I need to find A , k and B

A= squarerot of a^2 +b^2

tan(c) = b / a

Sin^2v = -1/2 I think

chat gpt says it so but im trying to figure out how it calculated it

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[regal tree]

I have a very simple question but somehow I cant come up with the correct interpretation myself

[regal tree]

The first equation should be maximized under the restrictions below

Which gives us this graph. In the graph, a downwards movement from (1) is already shown. Before the downwards movement, E4 is the optimum. After the downward movement the optimum is E5.

My question: How can we know that after a movement of (1) the optimum will still be where (1) and (2) cross? It seems quite intuitive that the new solution is bound by the same restrictions as the old solution, but I am looking for an explanation that makes me understand it better.

[vital surge]

consider 30x1 + 20x2 = c; imagine this as a line

geometrically this line has a slope that is steeper than (2) but less steep than (1)

as u increase the value of c, the line shifts out farther from the origin... it is no longer able to be shifted once it reaches the intersection point of (1) and (2)

hence maximized value of c occurs at that intersection

(may be worth noting this is only true if b is geq 80... if b less than 80 then (2) line is x = 0)

[regal tree]

The conclusion of the example where Im drawing this stuff from is exactly what you wrote down below: That the solution stays the same in a qualitative sense (intersection of 1 and 2) as long as 80 < b < 120.

Can the following assumption be made then?: If the initial solution is the crossing of (1) and (2), any new solutions after a movement of the restrictions will still be the crossing of (1) and (2), as long as there is an intersection that is valid.

[final saddle]

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[pine seal]

I'm currently doing a math exercise and I have the answr but I'm just confused on how to get it there.

[pine seal]

.status

!status

[final saddle]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[pine seal]

.close

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[west berry]

What's the question

[tranquil pine]

that’s sin(x) + something?

[tropic root]

to the function f(x) there is a maximum point and two inflection points in the domain.
the graph of function f'(x) is intercepting the x axis at point C. The graph of function f ''(x) is intercepting the x axis at points a and b. ( a< c<b) in this domain.
its known that f(a) - f'(a) = 0.263. and also the area between f'(x) , f''(x), the line x=a and the line x=c is 0.737.
Show that f(c) = 1.

well also f'(0) = f'(pi) = 0

f'(x) being Derivative and f''(x) being second Derivative

[final saddle]

@tropic root Has your question been resolved?

[tropic root]

<@&286206848099549185>

[zenith pollen]

it'll really help to draw a picture of f' and f'' where you get the x-intercepts and sign correct

then if you write out the integral of f'-f'' between a and c you can just integrate easily into f-f' |x=a to c and just plug in what you're given to get f(c)=1

[tropic root]

ok i tried to draw them let me know what u think

the x1 and 2 being a b c

[zenith pollen]

👍 that works, now if you draw the area between f' and f'' you'll be able to set up the integral and solve

[tropic root]

what does this help me with:
f(a) - f'(a) = 0.263?

[zenith pollen]

that'll come from the area integral, it'll turn into 4 terms and that piece of info let's you simplify 2 of them

like integral of f'-f'' from a to c

(since you see f' is above f'' over that interval)

[tropic root]

i cant see how the area is expressed in the graph

where are we talking?

[zenith pollen]

this part

the red dots are a and c

[tropic root]

ohh

i was thinking like X1 = a and c both cases

nvm

there is no way that was possible

yeah that makes sense

to calculate this i need to show both work from the positive part and the negativ part?

like how can i show f'(c) = 1

[zenith pollen]

do you have the area integral yet

like when you work through it f(c), f(a), and f'(a) will show up

[tropic root]

one sec il do it rq

oh ok got it

now it connects to the next section

i have which they say

and then i can find f(x)

that i can do i think

anyways thanks for your help

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[astral cedar]

for this problem, you need to do U-Sub, so set u=1+cosˆ2(x)
and du would be -sin2x since thats the property rule.
At this point you should have -1∫1/u * du
antiderivative of 1/u is just ln|u|+C. So you have -ln|u|+c, and now you just substitute it back in for -ln(|1+cosˆ2(x)|)+C
just read it, and doenst seem like you need help with it so i js wasted my time. whoops

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[waxen jay]

how can i convert the blue equation to cartisian?

[waxen jay]

so far i just multiplied both sides by 3-4cos(theta) but i cant seem to isolate the r and cosine theta to be by itself

[native portal]

use r = sqrt(x^2 + y^2) and x = r cos(theta)

[waxen jay]

is that all i have to do?

since this in terms of x or y

[native portal]

yea

you can remove the sqrt if u want

by squaring

[waxen jay]

alright thank you

🙏

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[charred tapir]

I normally understand how to solve systems of linear equations in three variables
this one confused me though:
1/x - 2/y + 3/z = -3
2/x - 3/y - 1/z = 7
3/x + 1/y - 2/z = 6
I know you need to turn this into a triangle system, but how to i simplify the fractions? I just need to understand that process

[hollow iron]

Say x' = 1/x, y' = 1/y and z' = 1/z

Then your system is linear in terms of x', y' and z'.

For example, the first equation turns to x' - 2y' + 3z' = -3

[charred tapir]

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[unique glacier]

does it make sense to talk about coordinates of some vector w in the proper subspace W of V with respect to some basis of the space V

[unique glacier]

for example W is a proper subspace of R^3 spanned by the vectors (1, 1, 0) and (0,0, 1)

w = 3(1,1,0) + 4(0,0,1)

and now i use the standart basis to say that the coordinates of w with respect to the standart basis of R^3 are (3, 3, 4)

idk if what i asked makes sense

[desert mantle]

well its still a vector in the big space

and there it makes complete sense to talk about the coordinates with respect to a basis of the whole space

[unique glacier]

ah ok good

thank you

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[gray acorn]

why is it wrong if

at the end it gives this

[cursive bough]

im not sure I understand the question

[gray acorn]

like i kinda just wanna know what i did wrong

on the paper above

[cursive bough]

is I just your shorthand for ln(x)/x^2?

[gray acorn]

yes

[cursive bough]

you need to evaluate the stuff on the right from 1 to 2

you shouldn't have an x anywhere

[gray acorn]

huh

im doing IBP

[gray acorn]

ah so

seperately?

[cursive bough]

or both at once is fine too

but you need to evaluate them at your bounds

[gray acorn]

yeah but

theyre both the same but different sign

so shouldnt it just be 0?

whatave the x is

[cursive bough]

oh I see

lemme look again

[gray acorn]

yeah thats where im stuck

[cursive bough]

oh I see

I thought I was going crazy for a sec lol

[cursive bough]

so here

you should have a 10 multiplied everywhere on the right too

you only multiply the integral by it

[gray acorn]

you mean divided?

[cursive bough]

so what you're doing just gives you that 10 times the integral is 10 times the integral

which like duh

not very helpful

[cursive bough]

no, multiplied

the integration by parts you did does not include the factor of 10

[gray acorn]

ah

well yk how the

this apart is the starting part

thats why at the end i put a 9

cuz it minus the 10

i moved it to the other side of the equation

[cursive bough]

you should have a 10 multiplied by everything on the right

[gray acorn]

wit but even if i do multiply

answer will still be 0

[cursive bough]

nope, if you multiply it will say 10*the integral = 10*the integral

which is obviously true

[gray acorn]

uhm

this is original quesiton btw

i just moved the 10 out

[cursive bough]

I know

[gray acorn]

ah so i multiply by 10

right

but

if like

is same equation

just inverse sign

[cursive bough]

these two things circled are equal

[gray acorn]

times 10 will be the same no?

yes

[cursive bough]

this is what the integration by parts does

if you look at the bottom circle thing, what does that simplify to

[gray acorn]

the same integration we started with

[cursive bough]

yes

so doing that method doesn't work is all that's telling you

[gray acorn]

ah so i cant do ibp or?

[cursive bough]

you can, just stop here

[gray acorn]

ohh

wait true

okay tysm!

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[sturdy cloak]

how they get 8 and 6

[final tangle]

start with expressimg 1m in cm

[sturdy cloak]

yh nvr mind

im thick as ice

mb

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[stray phoenix]

hi, idk how to calculate the value of a bc I'm not given the radius of the mass or anything 😭 could anyone help? am i misunderstanding something?